1may be the answer is wrong or may be e100 =( 271/2)2 which i don't thnk is
and my answer is as stated by pritish [1]
find the maximum product(approximate) of positive real numbers whose sum is 271 and also prove the result.
try it very good question and very pretty answer...
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39Let the real numbers be x and y.
We are given that
x + y = 271.
We have to maximise xy
Let xy = p
=> x(271 - x) = p
Differentiate wrt x.
271 - 2x = dp/dx
Put dp/dx = 0.
271 = 2x
=> x = 271/2
So y = 271/2.
Test by the double derivative.
d²p/dx² = -2 < 0
This implies x = 271/2 is a point of maxima.
So our real numbers are 271/2, 271/2.
This can also be proved by an axiom in algebra which states that -:
"When the sum of positive numbers is given to be constant, the product of these numbers is maximum when all the numbers are equal"
So your product is (271/2)² = 18360.25
1Pritish that is not the answer man
answer is e100
can anyone explain?????????how
62Lets assume that this is made of n numbers
then the maximum value will be when all of them are equal..
This part would be eas;y to prove using AM GM inequality
Hence the number would be (271/n)^n
So the aim now is to maximize (271/n)^n when n is an integer..
Anyone who wants to fill in the gaps from the proof?
1it is like this
USing inequaliy theorem for maxima and minima
i.e ,
IF a1 + a2 + ............... + an = k (constant )
for the value of a1.a2........an to be maximum a1 = a2 = ......... = an and maximum value of a1.a2........an is \left(\frac{k}{n} \right)^{n}
so we got here
a1 + a2 = 271
hence for maximum product we have a1.a2 = \left(\frac{271}{2} \right)^{2}
hence the answer ...........
162manmay, you have taken n =2
Why?
I mean after a step you have missed out the fact that n is not equal to 2
1bhaiya frm the inequality theorem na
as i stated which was given in the book
62\\\frac{d}{dn}\left( \frac{271}{n}\right)^n=\left( \frac{271}{n}\right)^nln(271/en)
Which is zero when 271=en
So 271/e =n
But Here what we have to keep in mind is that n is an integer..
so n = 100 will be the most appropriate integer closest to 271/e
So the maxima will be 2.71100
1i still don't get it
using inequality theorem as i stated abv in post #6
maximum value is \left(\frac{k}{n} \right)^{n} itself so why we further maximise it NISHANT BHAIYA ??
62yup gallardo i realised it at home after leaving office :P
Yesterday was "ALL India Bengal Band" called by CPI .. so couldnt come to office :D
at manmay.. no that is the value.. we have to maximize it.
It has not been menhtioned anywhere in the question that n=2
We have to maximize this given value over all values of n!
(What you are doing is that this is the maximum value for a fixed value of n...)
After that even n can vary
341Two links from the past (in another Avatar!)
http://www.goiit.com/posts/list/algebra-write-271-as-the-sum-of-positive-real-numbers-so-as-73541.htm#364087
and
http://www.goiit.com/posts/list/algebra-a-problem-in-maximization-everyone-s-invited-94553.htm
1thnk u NISHANT BHAIYA............. I FINALLY UNDERSTOOD AND GOT MY MISTAKE [1]